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Alexxx [7]
1 year ago
10

A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its no

se into the snow to catch its meal. If a fox jumps up to a height of 85 cm , calculate the speed at which the fox leaves the snow and the amount of time the fox is in the air. Ignore air resistance.

Physics
1 answer:
Pachacha [2.7K]1 year ago
4 0

Answer:

v = 4.08m/s₂

Explanation:

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the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of
nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
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2 years ago
5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by
Anna007 [38]

Answer:

a) W=2.425kJ

b) \Delta E=2.425kJ

c) T_f=20.06^{o}C

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

W=Fd

Where:

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F= force applied [N]

d= distance [m]

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d=h

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W=mgh

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which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

\Delta E=2.425kJ

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

\Delta Q=mC_{p}(T_{f}-T_{0})

Where:

Q= heat transferred

m=mass

C_{p}=specific heat

T_{f}= Final temperature.

T_{0}= initial temperature.

So we can solve the forula for the final temperature so we get:

T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}

So now we can substitute the data we know:

T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C

Which yields:

T_{f}=20.06^{o}C

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

\Delta Q=-2.425kJ

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2 years ago
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7 0
1 year ago
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Therefore, the work done is
W=Fd=(5 N)(2 m)=10 J
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2 years ago
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This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:

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2 years ago
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